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INCOMPRESSIBLE TWO PHASE FLOWS -THEORY
AND NUMERICS
Michael Ndjinga, Thi-Phuong-Kieu Nguyen, Christophe Chalons
To cite this version:
Michael Ndjinga, Thi-Phuong-Kieu Nguyen, Christophe Chalons. A 2 x 2 HYPERBOLIC SYSTEM
MODELLING INCOMPRESSIBLE TWO PHASE FLOWS -THEORY AND NUMERICS . 2016.
�hal-01303133�
TWO PHASE FLOWS - THEORY AND NUMERICS
MICHAEL NDJINGA †, THI-PHUONG-KIEU NGUYEN ‡, AND CHRISTOPHE CHALONS §
Abstract. We propose a 2 × 2 hyperbolic system of conservation laws to model the dynamics of two incompressible fluids in mechanical disequilibrium. In the theoretical part of the paper we show that this 1D system is not strictly hyperbolic, that the characteristic speed can not a priori be ordered and that the characteristic fields are neither genuinely nonlinear, nor linearly degenerate. We nevertheless prove the existence and uniqueness of an admissible solution to the Riemann problem. This solution remains bounded with positive volume fractions even when one the phases vanishes. We conclude that the multiphase/single phase transition does not imply mechanical equilibrium but displays a non classical wave structure.
In the numerical part of the paper we propose some approximate Riemann solvers to simulate the model, especially the multiphase/single phase transition. The classical Riemann solvers have been considered as Godunov scheme, Roe scheme with or without entropy fix. We also propose an in-cell discontinuous reconstruction method which proves to be successful, whereas the other schemes may show some spurious oscillations in some Riemann problem. Finally, as an application we study and simulate the problem of phase separation by gravity.
Key words. Riemann problem, incompressible fluid, hyperbolic system, genuinely non linear, two-fluid model.
1. Introduction The flow regime involved in a nuclear reactor core is purely liquid in normal operating conditions but may become a liquid-gas mixture in inciden-tal conditions or even purely gaseous in the case of a serious accident involving a core dewatering. The simulation of the single phase/multiphase transition is numerically challenging and has been a major difficulty in the design of new simulation platforms based on advanced two-fluid models, see for instance [11, 21] and references therein. An important issue is to guarantee the positivity of the volume fraction of each phase. There is an open debate as to whether this positivity is intrinsic to the conservation laws or requires some adequate source terms such as inter-phase friction. The ther-mal hydraulics platform CATHARE [12] assumes that when a phase disappears, its velocity is equal to the velocity of the other phase. In order to strongly couple the two phase velocities, they use a very high interfacial friction term to deal numeri-cally with these transitions. This paper intends to prove in the case of incompressible phases that the Riemann problem admits a positive solution without any frictional term and that the two velocities are not necessarily equal (section 2), to propose some numerical methods able to deal with vanishing phases (section 3) and then presents some numerical results (section 4). Allowing phase disappearance with mechanical disequilibrium enables for example the modelling of bubbles ascending in a liquid as a consequence of Archimedes’ principle.
1.1. The compressible model We consider a one dimensional isentropic two phase flow involving two fluids with densities ρ1< ρ2, pressures P1 and P2, volume
∗
†Den-Service de thermo-hydraulique et de m´ecanique des fluides (STMF), CEA, Universit´e
Paris-Saclay, F-91191 Gif-sur-Yvette, France, (michael.ndjinga@cea.fr).
‡Den-Service de thermo-hydraulique et de m´ecanique des fluides (STMF), CEA, Universit´e
Paris-Saclay, F-91191 Gif-sur-Yvette, France; LMV, UMR 8100, UVSQ, 45 avenue des ´Etats-Unis 78035 Versailles Cedex, France, (thi-phuong-kieu.nguyen@ens.uvsq.fr )
§LMV, UMR 8100, UVSQ, 45 avenue des Etats-Unis 78035 Versailles Cedex,´ France,
(christophe.chalons@uvsq.fr)
fractions α1∈ [0,1], α2= 1 − α1, and velocities u1 and u2. After averaging the mass
and momentum balance equations for each phase (see [7, 8, 6]), and neglecting mass and momentum transfer terms, the two-fluid model consists in the following four equations (k = 1,2) ∂tα1ρ1+ ∂x(α1ρ1u1) = 0, ∂tα2ρ2+ ∂x(α2ρ2u2) = 0, ∂t(α1ρ1u1) + ∂x(α1ρ1u21) + α1∂xP1= α1ρ1g, ∂t(α2ρ2u2) + ∂x(α2ρ2u22) + α2∂xP2= α2ρ2g. (1.1a) (1.1b) (1.1c) (1.1d) where g is the gravitational acceleration. Unlike [2, 3, 21, 11] we do not assume pressure equality P1= P2 nor do we introduce an interfacial pressure default term
4p∂xαk in the governing equations (1.1c) and (1.1d). Instead we consider a non zero
pressure difference of the form P1− P2=2(ρρ11−ρρ22)(u1− u2)
2 which yields a hyperbolic
system, see below. This pressure gap corresponds to a dynamic surface tension model accounting for the fact that velocity shear yields an increase of the microscale interfa-cial curvature via the well known Kelvin-Helmholtz instability (see[10]). Taking into account surface tension, the increase of local curvature results in a pressure difference via the Laplace law P1− P2= γσ which should vanish only when u1= u2. The kinetic
energy gap 1 2ρ1u 2 1−12ρ2u 2 2=12 (ρ1u1−ρ2u2)2 ρ1−ρ2 − 1 2 ρ1ρ2(u1−u2)2
ρ1−ρ2 is related to the
momen-tum gap ρ1u1− ρ2u2and to the velocity gap u1− u2. In this first study, we make the
simple assumption that the pressure gap exactly compensates the contribution of the velocity gap to the kinetic energy gap. The resulting model is in fact a minimal per-turbation of the B.L Keyfitz [2] using a pressure difference proportional to (u1− u2)2
which yields the hyperbolicity.
The system (1.1a-1.1d) has four main unknowns: α1,P1,u1,u2, the other unknowns
can be obtained using the equations of state ρk(Pk) and the pressure gap law
P1− P2=
ρ1ρ2
2(ρ1− ρ2)
(u1− u2)2.
Defining the mixture sound wave cm=
q(α
1ρ2+α2ρ1)c22c21
α1ρ2c22+α2ρ1c21
, where ck= ck(Pk),k = 1,2
are the sound speeds of each phase, following the work in [4], the Taylor expansion of the four eigenvalues of the system (1.1a-1.1d) when u1− u2 cmis
λ1= ρ1u1− ρ2u2 ρ1− ρ2 1 − ρ1ρ2 (α1ρ2+ α2ρ1)2 + O u1− u2 cm
void fraction wave,
λ2= ρ1u1− ρ2u2 ρ1− ρ2 + O u1− u2 cm
void fraction wave,
λ3= α1ρ2u1+ α2ρ1u2 α1ρ2+ α2ρ1 + cm+ O u1− u2 cm acoustic wave, λ4= α1ρ2u1+ α2ρ1u2 α1ρ2+ α2ρ1 − cm+ O u1− u2 cm acoustic wave.
Thus for small relative velocities u1− u2 cm, the system (1.1a-1.1d) is hyperbolic
with 2 acoustic waves involving the mixture sound speed and two void waves that are specific to the two phase dynamics. Since we are interested by the void wave dynamics
and flows at low Mach numbers, we devote more attention to the incompressible limit of the system (1.1a-1.1d).
1.2. The incompressible model In order to study more precisely the volume fraction waves involved in applications where the fluid densities are almost constant, we follow [2] and assume that both phases are incompressible with constant densities ρ1and ρ2.
First remark that suming ρ1
1(1.1a) and
1
ρ2(1.1b) yields ∂x(α1u1+ α2u2) = 0, and
therefore the value of K = α1u1+ α2u2 may be determined from the boundary
condi-tions. From now on, we assume that K is a constant both in space and time for sim-plicity. For simplification, we will assume that K = 0, which is true for example if there is a wall boundary condition. If K 6= 0 is constant both in time and space, a Galilean change of reference frame u0k= uk− K yields a new system with K0= α1u01+ α2u02= 0
(see [2]). It is then possible obtain a close system of two equations by setting α = α1,
ω = ρ1u1− ρ2u2.
. (1.2)
Note that the velocities u1 and u2can be recovered from (1.2) and K = 0 as
u1= (1 − α)ω α(ρ2− ρ1) + ρ1 + K, u2= −αω α(ρ2− ρ1) + ρ1 + K. (1.3)
The unknown vector U =t(α,ω) can be described by a conservative system
∂tU + ∂xF (U ) = S, (1.4) with F (U ) = α(1−α)ω α(ρ2−ρ1)+ρ1 ω2 2(ρ1−ρ2) , S = 0 (ρ1− ρ2)g ! , (1.5) ∇F (U ) = ω ρ1−ρ2 1 − ρ1ρ2 (α(ρ2−ρ1)+ρ1)2 α(1−α) α(ρ2−ρ1)+ρ1 0 ρ ω 1−ρ2 . (1.6)
The first equation of (1.4) is obtained fromρ1
1(1.1a) and the second of (1.4) comes from
the linear combination of equations 1
α1(1.1c) −
1
α2(1.1d) (see the details in Appendix
5.1).
2. Theoretical study
2.1. Hyperbolicity and characteristic fields The following theorem is a direct consequence of formula (1.6).
Theorem 2.1 (Hyperbolicity of system (1.4)). The jacobian matrix (1.6) of the system (1.4) always admits real eigenvalues
λ1= ω ρ1− ρ2 1 − ρ1ρ2 (α(ρ2− ρ1) + ρ1)2 , λ2= ω ρ1− ρ2 , (2.1) provided U ∈ H where H = {(α,ω),α ∈ [0,1],ω ∈ R}.
α ω α = 0 α = 1 ω = 0 H+ H−
Fig. 2.1: Strictly hyperbolic domain H∗= H+∪ H−∪ {(0,ω),ω ∈ R} ∪ {(1,ω),ω ∈ R}.
Morever, such a matrix is diagonalisable provided (α,ω) ∈ H∗ where
H∗= H+∪ H−∪ {(0,ω),ω ∈ R} ∪ {(1,ω),ω ∈ R},
and H±=(α,ω),ω ∈ R∗±, and α ∈ (0,1) .
In general, the system (1.4) is weakly hyperoblic on the domain H = H∗∪ {(α,0),α ∈
(0,1)}.
The states (ω = 0,α = 0) and (ω = 0,α = 1) will play an important role in connecting states in H+to states in H− and will be called critical states.
H is neither an open nor a simply connected subset of R2, see Figure 2.1. The
eigenvectors of ∇F are
~r1=t(1,0), ~r2=t(α(1 − α)(ρ1− ρ2)(α(ρ2− ρ1) + ρ1),ρ1ρ2ω). (2.2)
Assuming that ρ1< ρ2, we notice that the two eigenvalues are not a priori ordered
since λ1< λ2 if ω < 0, λ1> λ2 if ω > 0, λ1= λ2 if ω = 0. (2.3)
Moreover, the signs of ~∇λ1·~r1=(α(ρ−2ρ2−ρ11ρ)+ρ2ω1)3 and ~∇λ2·~r2=(ρρ11−ρρ2ω2) are not known
a priori since ~ ∇λk·~rk> 0 if ω < 0, ~ ∇λk·~rk< 0 if ω > 0, ~ ∇λk·~rk= 0 if ω = 0, (2.4)
where k = 1,2. Therefore the characteristic fields associated to λ1and λ2are genuinely
nonlinear in each domain H+ and H−, but are neither genuinely non linear, nor
linearly degenerate in general.
2.2. Triangular systems of conservation laws The system (1.4) without source terms is a particular case of triangular systems of hyperbolic conservation laws due to the fact that the second equation is independent of the first one. Let us consider the simplest triangular system of conservation laws
∂tα + ∂xg(α,ω) = 0, (2.5a)
which has been studied largely in the literature. For example, the reader is referred to [28, 23, 24, 26, 27] and references therein for more details. Inspired by the theory of scalar conservation laws, an simple way to solve a triangular system of 2 × 2 equations is to compute the solution of the independent equation then replace it to the remaining one. However for the general results of existence and uniqueness to the Cauchy problem (or even the Riemann problem) for the system (2.5a-2.5b) with regular functions of f (ω) and g(α,ω) is still open. The difficulties encountered include the non strict hyperbolicity and the resonance.
Let us introduce well-known results for the triangular system as well as the ap-proaches so that we have a general point of view and can figure out our contribution in this interest. The first approach is to generalize the weak solutions, i.e. to extend the space of an admissible solution in the sense that they are not necessarily bounded. Following this approach, some authors have been studying non strictly hyperbolic triangular systems. More precisely, considering a particular system of (2.5a-2.5b) where g(α,ω) = αω and f (ω) = ω2, the authors in [28, 23] introduced the
class of admissible solutions to the Riemann problems admitting delta-shocks. Such study has been extended by many researchers, for example in [24, 25] and references therein. The general results obtained for the Cauchy problem in such references require the function g(α,ω) to be linear with respect to α for each ω, i.e. g(α,ω) can be rewritten as g(α,ω) = h(ω)α. The key idea in this case is to use the study of the linear transport equation where the velocity admits the discontinuities in space and in time. The resulting admissible solutions include the delta-shock waves, which appear around the configuration of non strictly hyperbolic.
Another approach can be found in [26, 27], where the authors used the theory of compensated compactness to prove the existence and uniqueness of the Cauchy problem for a triangular hyperbolic system. The main idea is to use numerical schemes, as Godunov-type in [26] or the relaxation scheme in [27], in order to prove the convergence of the schemes which implies the existence of the solution. However, the proof of the convergence strongly depends on the assumption of the function g(α,ω). More precisely, the function g(α,ω) must be genuinely nonlinear in α for any ω in the interesting domain, i.e. ∀ω, gαα(α,ω) 6= 0 .
We can not apply directly the results in the references presented above to the system (1.4) although there are some similar properties between our system and the one in [28, 23], such as not strict hyperbolicity when ω = 0, the two eigenvalues are not ordered, etc. The reason is that our flux function in equation (2.5a) is not linear with respect to α, therefore we are not able to use directly the theory of transport equations. Moreover, the function g(α,ω) is not genuinely nonlinear with respect to α. We then present in this documenta new approach yielding the existence and uniqueness of the admissible solution to the Riemann problem for the system (1.4) without source terms. We start by defining the admissible solutions.
2.3. Admissible solutions of the Riemann problem The fact that the domain H is not open, that the system is not strictly hyperbolic and that the charac-teristic fields are neither genuinely nonlinear neither linearly degenerate raises many theoretical as well as numerical difficulties. We cannot use the classical Lax theorem (see [19]) to obtain solutions to the Riemann problem but will however build solutions to the Riemann problem having a non classical wave structure for any pair of left and right states data in H.
Definition 2.1 (Hugoniot locus and Riemann invariants). Given a state U ∈ H, the Hugoniot locus S(U ) associated to (2.12) and U is the set of states that can be connected to U via a shock wave:
S(U ) = {V ∈ H, ∃σ(U,V ) ∈ R, F (U ) − F (V ) = σ(U,V )(U − V )}.
For any k ∈ {1,2}, a k-Riemann invariant associated to (2.12) is a function Rkdefined
on H such that
∀U ∈ H, ∇Rk·~rk= 0.
and the k-rarefaction wave associated to the Riemann invariant Rk at a left state U
is
k-R(U ) = {V ∈ H, Rk(V ) = Rk(U ), λk(V ) ≥ λk(U )}.
We did not find an entropy to our system and instead of an entropy criterion we use the Liu criterion to define admissible solutions.
Definition 2.2 (Admissible solution of the Riemann problem). An admissible solution to (2.12) is a weak solution that is composed of a finite number of constant states connected by a rarefaction wave or a shock wave, connecting left and right sates UL and UR and propagating at a speed σ, that satisfies the Liu criterion:
σ(UL,UR) ≤ σ(UL,U ), ∀U ∈ S(UL) between UL and UR.
2.4. Shock and rarefaction waves
In the theory of 2 × 2 strictly hyperbolic systems with genuinely nonlinear char-acteristic fields, each eigenvector family ~rk, k = 1,2 defines at any state U0one single
shock and one single rarefaction curve, both starting and having tangent vector ~rk
at U = U0. These two curves are defined in an open neighborhood of U0 and do not
necessarily extend to the entire domain, see [18], [19] for example.
Our system is neither strictly hyperbolic neither genuinely nonlinear, our strictly hy-perbolic domain is not open. However we prove below that there are two families of shock “curves” (1-shocks and 2-shocks) and rarefaction “curves” (1-rarefactions and 2-rarefactions) and that they extend to the entire domain.
The originality in our system is the fact that the shock “curve” associated to a state is not necessarily a connected set. Indeed the 1-shock “curve” associated to the states (α ∈ [0,1],ω = 0) is a contact discontinuity whose speed is σ = 0, whereas there is no 1-rarefaction associated to such states. Moreover the 2-shock family of “curves” can be composed of the two branches of a hyperbola and there are more than one 2-rarefaction curves passing through the critical state (α = 1,ω = 0).
We now characterise the shock and rarefaction curves and illustrate them on Figure 2.2. We note that the entropy criterion is not taken into account in the following theorem.
Theorem 2.2 (Existence of shock curves).
Any state U0= (α0,ω0) ∈ H \ (α = 1,ω = 0) belongs to two shock curves in H. The state
(α = 1,ω = 0) belong to three shock curves in H.
The equation of the 1-shock family (1-S) of states (α,ω1(α)) connected to the state
U0= (α0,ω0) is
• If ω0= 0, ω1(α) = 0, the 1-shock ”curve” degenerates a contact discontinuity
whose the speed is σ = 0.
The equation of the 2-shock family (2-S) of states (α,ω2(α)) connected to the state
U0= (α0,ω0) is • If ω06= 0 and α0∈ (0,1) ω2(α) = α(ρ2− ρ1) + ρ1 α0(ρ2− ρ1) + ρ1 × α 2 0(ρ2− ρ1) + α0(ρ1− 2ρ2+ αρ2− αρ1) + αρ1 α2(ρ 2− ρ1) + α(ρ1− 2ρ2+ α0ρ2− α0ρ1) + α0ρ1 ω0
• If ω06= 0 and α0= 0, then either α = 0 or ω2=
(α(ρ2−ρ1)+ρ1)ω0
α(ρ2−ρ1)+ρ1−2ρ2.
• If ω06= 0 and α0= 1, then either α = 1 or ω2=−(α(ρ−α(ρ2−ρ1)+ρ1)ω0
2−ρ1)+ρ1 .
• If ω0= 0 and α0∈ [0,1), the 2-shock curve is straight line defined by the
con-stant α which is the unique solution in [0,1) of the following equation α2(ρ2− ρ1) + α(ρ1− 2ρ2+ α0ρ2− α0ρ1) + α0ρ1= 0.
• If ω0= 0 and α0= 1, then either α = 1 or α =ρρ1
2−ρ1 if ρ2> 2ρ1.
Proof: A state U in S(U0) must fulfil the Rankine-Hugoniot condition
F (U ) − F (U0) = σ(U,U0)(U − U0), (2.6)
which for the system (2.12) takes the form ( α(1−α)ω α(ρ2−ρ1)+ρ1− α0(1−α0)ω0 α0(ρ2−ρ1)+ρ1= σ(U,U0)(α − α0), ω2 2(ρ1−ρ2)− ω2 0 2(ρ1−ρ2)= σ(U,U0)(ω − ω0). (2.7)
• If ω = ω0, there always exists σ(U,U0) such that (2.7). This comes from the
fact that the second equation is independent from α. The 1-shock family is made of states sharing the same values of ω which is consistent with the expression of the eigenvector ~r1 (equation 2.2).
• If ω 6= ω0, the second equation of (2.7) implies σ(U,U0) =2(ρω+ω1−ρ02), which yields
in the first equation α2(ρ 2− ρ1) + α(ρ1− 2ρ2+ α0ρ2− α0ρ1) + α0ρ1 α(ρ2− ρ1) + ρ1 ω −α 2 0(ρ2− ρ1) + α0(ρ1− 2ρ2+ αρ2− αρ1) + αρ1 α0(ρ2− ρ1) + ρ1 ω0= 0. (2.8) – If ω0= 0, (2.8) implies α2(ρ2− ρ1) + α(ρ1− 2ρ2+ α0ρ2− α0ρ1) + α0ρ1= 0. (2.9)
It is easy to see that ∀α ∈ [0,1), the equation (2.9) has a unique solution α ∈ [0,1), and such an α defines the straight line 2-shock curve connect-ing to (α0,ω0= 0). Moreover, the state (α0= 1,ω0= 0) connects to the
straight line α = 1 and can connect to all states (α = ρ1
ρ2−ρ1,ω ∈ R
∗) under
the assumption that ρ2> 2ρ1.
– If ω06= 0 and α0= 0, then (2.8) implies either α = 0 or ω = (α(ρ2−ρ1)+ρ1)ω0
– If ω06= 0 and α0= 1, then (2.8) implies either α = 1 or ω = −(α(ρ2−ρ1)+ρ1)ω0
−α(ρ2−ρ1)+ρ1 .
– If ω06= 0 and α0∈ (0,1), then then (2.8) implies
ω(α) = α(ρ2− ρ1) + ρ1 α0(ρ2− ρ1) + ρ1 × α 2 0(ρ2− ρ1) + α0(ρ1− 2ρ2+ αρ2− αρ1) + αρ1 α2(ρ 2− ρ1) + α(ρ1− 2ρ2+ α0ρ2− α0ρ1) + α0ρ1 ω0.
The shape of the shock curves can be seen on Figure 2.2.
2 Theorem 2.3 (Existence of rarefaction curves).
• Any state U0= (α0,ω0) ∈ H \ {(α,ω = 0)} belongs to two rarefaction curves,
the 1-rarefaction curve (1-R) is described by the equation ω1(α) = ω0 and the
2-rarefaction curve (2-R) is described by the equation
ω2(α) = ω0 α(ρ2− ρ1) + ρ1 α0(ρ2− ρ1) + ρ1 α α0 ρ1−ρ2ρ2 1 − α 1 − α0 ρ1−ρ2−ρ1 (2.10)
• The state (α0= 0,ω0= 0) belongs to a single rarefaction curve, the
2-rarefaction curve described by α = 0.
• The state (α0= 1,ω0= 0) belongs to all 2-rarefaction curves going through
U0= (α0,ω0) ∈ H \ {(α = 0,ω = 0)} and described by (2.10) if α06= 1 or α = 1,
otherwise.
• There is no rarefaction going through the state (α0∈ (0,1),ω = 0).
Proof: From the definition of Riemann invariants, R1 and R2 must satisfy
∂R1 ∂α = 0; α(1 − α)(ρ1− ρ2)(α(ρ2− ρ1) + ρ1) ∂R2 ∂α + ρ1ρ2ω ∂R2 ∂ω = 0.
Since R1 is function of only ω, it is easy to obtain the equation of the 1-rarefaction
curves.
Considering the Riemann invariant R2,
• if α ∈ (0,1), we have ∂ω ∂α= ρ1ρ2ω α(1 − α)(ρ1− ρ2)(α(ρ2− ρ1) + ρ1) . (2.11)
Solving the linear ordinary differential equation (2.11), we obtain explicitly the equation of the rarefaction curves (2.10) in the theorem.
• if α ∈ {0,1}, then – either ∂R2
∂ω = 0, the rarefaction curves are α(ω) = 0 (or α(ω) = 1)
corre-sponding to α0= 0 (or α0= 1),
– or ω = 0. However, only the state (α = 1,ω = 0) belongs to all R2
described by (2.10) due to continuing property (limα→1ω2(α) = 0,
limα→0ω2(α) = ±∞).
2 The structure of the rarefaction curves is illustrated on Figure 2.2.
(a) 1-shock curves (b) 1-rarefaction curves
(c) 2-shock curves (d) 2-rarefaction curves
Fig. 2.2: Shock curves and Rarefaction curves. U0= (0,40): green. U0= (0.5,−10):
red. U0= (0.5,10): violet. U0= (1,−20): blue.
2.5. Solution of the Riemann problem We consider the Riemann problem for the conservative system (1.4) in the case g = 0 with a piecewise constant initial data: ∂tU + ∂xF (U ) = 0 U0(x) = UL(αL,ωL) if x ≤ 0, UR(αR,ωR) if x > 0, (2.12)
where U =t(α,ω) ∈ H, F (U ) are defined in (1.5), and U
L,UR∈ H. We start by stating
and proving in section 2.5.1 existence and uniqueness of an admissible solution to (2.12) in Theorem 2.5. We then give in section 2.5.2 four important examples of non classical solutions to the Riemann problem in order to illustrate the proof of Theorem 2.5 and the unusual behaviour of the system (2.12) as well.
2.5.1. Main result We intend to prove existence and uniqueness of an admis-sible solution. Due to the complex structure of the shock curves, there are numerous different cases to be considered. We first describe the structure of admissible solutions in the following lemma.
Lemma 2.4. Assume that an admissible solution of the Riemann problem contains two adjacent waves which connect the left state UL= (αL,ωL) ∈ H to the right state UR=
(αR,ωR) ∈ H through the intermediate state UI= (αI,ωI). Then, these two adjacent
waves must be in different family (i.e. one wave from a 1-family and the other from a 2-family) at the exception that a 2-rarefaction may be followed by a 2-rarefaction. Moreover, if the two adjacent waves are a 1-wave followed by a wave (resp. a 2-wave followed by a 1-2-wave), then the intermediate state satisfies ωI= ωL≤ 0 (resp.
ωI= ωR≥ 0). Therefore, the value of ω satisfies the maximum principle.
Proof: This lemma is a direct consequence of the characterisation of k-shock curves and k-rarefaction curves (theorems 2.2 and 2.3) and of the speed order criterion. Assume first that two adjacent waves are two 1-shock waves connecting the left state UL= (αL,ωL) to the right state UR= (αR,ωR) through the intermediate state UI=
(αI,ωI). A consequence of Theorem 2.2 is that ωL= ωR= ωI. From the first equation
in (2.7), the speed of the first 1-shock and the second 1-shock is
σ11= ωL ρ1− ρ2 1 − ρ1ρ2 (αI(ρ2− ρ1) + ρ1)(αL(ρ2− ρ1) + ρ1) , and (2.13) σ12= ωR ρ1− ρ2 1 − ρ1ρ2 (αI(ρ2− ρ1) + ρ1)(αR(ρ2− ρ1) + ρ1) . (2.14)
Liu’s criterion for both shocks gives αR> αI> αL if ωL= ωR= ωI> 0 and αR< αI<
αL if ωL= ωR= ωI< 0, while the speed order criterion σ11< σ12 gives αR< αL if
ωL= ωR= ωI> 0 and αR> αLif ωL= ωR= ωI< 0. Therefore, the admissible solutions
of the Riemann problem do not admit two adjacent waves in the same 1-shock family. The conclusion for the 1-rarefaction family and 2-shock family are completely the same while the conclusion for the 2-rarefaction family is an exception.
Since the two 2-rarefaction curves join at unique point (α = 1,ω = 0), if two adjacent waves are two 2-rarefaction, then ωL> 0 > ωR. Such two 2-rarefaction satisfy the
speed criterion although there exists only an intermediate point (α = 1,ω = 0) (no intermediate constant state). It is consistent with the conclusion of a system in [22]. As for the second statement of the theorem, assume that the intermediate state UI=
(αI,ωI) connects to the left state (resp. right state) by a 1-wave and connects to the
right state (resp. the left state) by a 2-wave, due to the fact that the equation of 1-wave is that ω is constant, see theorem of existence of shock curves Theorem 2.2 and rarefaction curves Theorem 2.3, then ωI= ωL(or ωI= ωR). In order to prove the
sign property of ωI= ωL in the case of a 1-wave followed by a 2-wave, let us notice
that for the solution to be admissible, the speed of propagation of the 1-wave must be smaller or equal to the one of the 2-wave. (2.3) thus imposes ωI= ωL≤ 0. Similarly,
we also obtain the sign property ωI= ωR≥ 0 in the case of a 2-wave followed by a
1-wave.
2 Theorem 2.5 (Existence and uniqueness for the Riemann problem). For any pair UL(αL,ωL),UR(αR,ωR) ∈ H, the Riemann problem admits a unique
admissi-ble solution U (x,t) ∈ H which depends continuously on αL and αR with some pair
(ωL,ωR) ∈ R2 .
Proof: Let UL= (αL,ωL) 6= UR= (αR,ωR) be in H. Assuming that ρ1< ρ2, we find
admissible solutions satisfying the initial data (2.12).
Case 1: ωL= ωR. The solution is a single 1-wave. In particular, if ωL= ωR= 0 and
αL6= αR, such a 1-wave is the degenerate 1-contact discontinuity.
Case 2: ωL> ωR. An admissible solution will contain at least one 2-rarefaction (see
Theorem 2.3 for the existence of the rarefaction curves which are illustrated in Figure 2.2). In detail, we consider two possibilities, ωLωR≥ 0 or ωLωR< 0.
• Case 2.1: ωLωR≥ 0. Without loss of generality, we assume that ωL and ωR
are non-positive. From (2.3), an admissible solution is a 2-rarefaction followed by a 1-wave. In addition, the 2-rarefaction monotonic curve ω2(α) intersects
the 1-wave curve, ω(α) = ωR, at a unique point, see also Figure 2.2. The
uniqueness of the admissible solution is thus obtained.
• Case 2.2: ωL> 0 > ωR. Due to Lemma 2.4, the left state must connect to a
2-rarefaction and the right state does the same (otherwise it violates Lemma 2.4). This property then implies the uniqueness of the admissible solution whose structure depends on the values of αL and αR, there are three
possi-bilities
– If αL6= 0 and αR6= 0, the admissible solution is a 2-rarefaction followed
by another 2-rarefaction. See Example 2.3 and Figures 2.4(a) and 2.4(c). In particular, if (αL= 1,ωL= 0) or (αR= 1,ωR= 0), the solution is a
single 2-rarefaction.
– If αL= αR= 0, the admissible solution is a single 2-rarefaction due to
existence of a 2-rarefaction which goes through α = 0, see Theorem 2.3. – If (αL= 0 and αR6= 0) or (αL6= 0 and αR= 0), due to Theorem 2.3 the
2-rarefaction connecting the left state and the one connecting the right state are neither coincide nor intersecting, the admissible therefore con-tains more than two waves. The unique result solution which satisfies the speed criterion is a 2-rarefaction attached to a 1-contact disconti-nuity and then followed by another 2-rarefaction. See Example 2.4 and Figures 2.4(b) and 2.4(d).
Case 3: ωL< ωR. This case is more technical and the uniqueness of the admissible
solution has to be carefully studied. An admissible solution in this case contains a 2-shock, since ω must increase from ωL to ωR, see Figure 2.2 for the shock and
rarefaction curves.
We introduce a new variable
β = α(ρ2− ρ1) + ρ1. (2.15)
Since α ∈ [0,1], we have β ∈ [ρ1,ρ2].
First of all, we look for admissible solutions whose structure is a 2-shock followed by a 1-wave with U∗ as intermediate state. The necessary condition ω∗= ωR≥ 0 follows
from Lemma 2.4. In order to select an admissible solution, we introduce the speed order criterion in this specific case, which is
σ2≤ λ1(U∗) if the solution is a 2-shock followed by a 1-rarefaction, (2.16)
or σ2< σ1 if the solution is a 2-shock followed by a 1-shock, (2.17)
where σ2=2(ρωL+ωR 1−ρ2), σ1= ωR ρ1−ρ2 1 − ρ1ρ2 β∗β R , U∗= (α∗,ωR) such that α∗=ρ1−β ∗ ρ1−ρ2. For
simplicity, we can rewrite the inequalities (2.16) as β∗≤q2ρ1ρ2ωR
ωR−ωL and (2.17) as β
∗≤ 2ρ1ρ2ωR
βR(ωR−ωL).
Replacing α by β−ρ1
ρ2−ρ1 in the 2-shock curve equation in Theorem 2.2 yields after some
calculations that all states (α,ωR) which are connected to UL by a 2-shock satisfy the
following quadratic equation
Let G(βL,β) be the left hand side of the equation (2.18), then
G(βL,ρ1) = −ρ1(βL− ρ1)(βL(ωR− ωL) + 2ρ2ωL) and (2.19)
G(βL,ρ2) = ρ2(ρ2− βL)(βL(ωR− ωL) + 2ρ1ωL). (2.20)
Recall that ωL< ωR, the concave quadratic function G(βL,β) may have no solution
or more than one solution. We look for a condition on ωR such that the
equa-tion (2.18) has a non-negative soluequa-tion α∗∈ [0,1], namely β∗∈ [ρ
1,ρ2], or equivalently
G(βL,ρ1)G(βL,ρ2) ≤ 0 because G(βL,ρ1) and G(βL,ρ2) can not be negative at the
same time. By considering a variation of αL, we get different cases.
• Case 3.1 If αL∈ {0,1}, then G(βL,ρ1)G(βL,ρ2) = 0 and the equation (2.18)
may have two solutions with β∗∈ [ρ
1,ρ2]. Using the speed order criterion, we
obtain:
– Case 3.1.1 If αL= 0, i.e. βL= ρ1 the two potential solutions of (2.18)
are β∗= ρ1and β∗=ω2ρR2−ωωRL. However ω2ρR2−ωωRL violates the criteria (2.16)
and (2.17). Therefore, only β∗= ρ1 is acceptable and the admissible
solution is a 2-shock followed by a 1-shock (not a 1-rarefaction) since ωI= ωR> 0 (see 1-shock curves and 1-rarefaction curves on Figure 2.2).
Such an admissible solution satisfies the criterion (2.17) if and only if βR≤ω2ρ2ωR
R−ωL.
– Case 3.1.2 If αL= 1 and βL≥−2ρωR−ω2ωLL, (which is equivalent to ωR≥ −ωL
and also to 2ρ1ωR
ωR−ωL≥ ρ1), then there exists a unique solution β
∗=
min{ρ2,ω2ρR1−ωωRL} satisfying the criteria (2.16) and (2.17).
• Case 3.2 If αL∈ (0,1), or equivalently βL∈ (ρ1,ρ2), then it is obvious that
G(βL,ρ1) and G(βL,ρ2) can not be non-positive at the same time, the
equa-tion (2.18) therefore has at most a soluequa-tion β∗∈ [ρ 1,ρ2].
The first possibility is G(βL,ρ1) ≤ 0 and G(βL,ρ2) ≥ 0, or equivalently
βL≥
−2ρ2ωL
ωR− ωL
, (2.21)
i.e. G(βL,β) ≥ 0 implies β ∈ [β∗,+∞).
The second possibility is G(βL,ρ1) ≥ 0 and G(βL,ρ2) ≤ 0, or equivalently
βL≤
−2ρ1ωL
ωR− ωL
. (2.22)
i.e. G(βL,β) ≥ 0 implies β ∈ (−∞,β∗].
As long as the condition (2.21) or (2.22) is satisfied, the speed order criterion (2.16) and (2.17) will help us to select an admissible solution.
– Case 3.2.1 Assume first that the admissible solution is a 2-shock followed by a 1-rarefaction and use the speed order criterion (2.16). We denote H(βL) = G βL, q 2ρ1ρ2ωR ωR−ωL
and first prove that H(βL) > 0 for all βL∈
(ρ1,ρ2). Writing H (βL) explicitly,
H(βL) = −4ρ1ρ2βLωR+ (βL(ωR− ωL)(2(ρ1+ ρ2) − βL) + 2ρ1ρ2ωL)
r 2ρ1ρ2ωR
ωR− ωL
Calculating and evaluating H(ρ1), H(ρ1) = −4ρ21ρ2ωR+ ρ1(ρ1(ωR− ωL) + 2ρ2ωR) r 2ρ1ρ2ωR ωR− ωL ≥ −4ρ2 1ρ2ωR+ 2ρ1 p 2ρ1ρ2ωR(ωR− ωL) r 2ρ1ρ2ωR ωR− ωL (2.23) = 0.
The inequality (2.23) is obtained by the Cauchy’s inequality for two non-negative numbers ρ1(ωR− ωL) and 2ρ2ωR. The result H(ρ2) ≥ 0
is obtained similarly. Moreover, considering βL as a variable of the
quadratic function H(βL) whose highest order’s coefficient is negative
and both H(ρ1) and H(ρ2) are non-negative, we achieve H(βL) > 0 for all
βL∈ (ρ1,ρ2). This result shows that the condition (2.21) satisfies the
cri-terion (2.16) (since GβL, q 2ρ1ρ2ωR ωR−ωL = H(βL) and G(βL,β) ≥ 0 implies β ∈ [β∗,+∞), so that β∗≤q2ρ1ρ2ωR
ωR−ωL) while the condition (2.22) is
im-possible to satisfy (since it implies β∗≥q2ρ1ρ2ωR
ωR−ωL because G(βL,β) ≥ 0
implies β ∈ (−∞,β∗] in this case).
– Case 3.2.2 Assume that the admissible solution is a 2-shock followed by a 1-shock and use the speed order criterion (2.17). If 2ρ1ρ2ωR
βR(ωR−ωL)>
ρ2, the criterion (2.17) is always satisfied. So we will merely consider 2ρ1ρ2ωR βR(ωR−ωL)≤ ρ2, where we define M (βL) = G βL,βR2ω(ωRRρ−ω1ρ2L) . Rewrite M (βL) as the following M (βL) = 2ρ1ρ2ωR β2 R(ωR− ωL) ((βRωL− βLωR) + βLβR(ωR− ωL)(2ρ1+ 2ρ2− βL− βR)). We compute M (ρ1) = 2ρ1ρ2ωR βR (ρ1− βR) ρ1− 2ρ1ρ2ωR βR(ωR− ωL) , M (ρ2) = 2ρ1ρ2ωR βR (ρ2− βR) ρ2− 2ρ1ρ2ωR βR(ωR− ωL) .
The condition (2.21) implies M (ρ1) ≥ 0 and M (ρ2) ≥ 0, and then
ob-viously M (βL) > 0, ∀βL∈ (ρ1,ρ2) (since the second order polynomial
βL→ M (βL) is concave), this result satisfies the criterion (2.17).
On the other hand, the condition (2.22) violates the criterion (2.17) as long as we assume the solution has more than one wave. By the conti-nuity, this solution is not admitted.
We summarize the structure of the admissible solution according to the initial data. • If αL∈ (0,1] and βL≥−2ρωR−ω2ωLL, the solution is a 2-shock followed by a 1-wave.
• If αL= 0 and βR≤ω2ρR2−ωωRL, the solution is a 2-shock followed by a 1-shock
(α∗= 0).
Similarly, we obtain the following results
• If αR= 0 and βL≤−2ρω 2ωL
R−ωL, the solution is a 1-shock followed by a 2-shock
(α∗= 0).
Before continuing the proof, we can conclude that if αL6= 0 and αR6= 0, the solution
consisting of a 2-shock (resp. 1-wave) followed by a 1-wave (resp. a 2-shock) is admissible if βL≥−2ρω 2ωL
R−ωL (resp. βR≥
2ρ2ωR
ωR−ωL).
The rest of our proof considers the initial data which are not studied above, i.e. αLαR6= 0 and βL<−2ρω 2ωL
R−ωL and βR<
2ρ2ωR
ωR−ωL. According to Lemma 2.4, an admissible
solution must be a 1-wave followed by a 2-shock connected to another 1-wave. Let us denote the two intermediate states ordered from the left to the right by U∗(α∗,ωL)
and U∗∗(α∗∗,ωR).
• If α∗6= 0 and α∗∗6= 0, due to the previous results a 1-wave followed by a
2-shock is admissible if β∗∗≥ 2ρ2ωR
ωR−ωL and this 2-shock followed by another
1-wave is admissible if β∗≥−2ρ2ωL ωR−ωL. Both β ∗∗≥ 2ρ2ωR ωR−ωL and β ∗≥−2ρ2ωL ωR−ωL are
satisfied if and only if ωL= −ωR and β∗= β∗∗= ρ2, this condition however
violates the speed order criterion.
• If (α∗6= 0 and α∗∗= 0) or (α∗= 0 and α∗∗6= 0), Theorem 2.2 shows that it is
impossible.
• α∗= α∗∗= 0 is admissible, this solution satisfies all criteria of speed order.
See Figure 2.3(b) for the construction of such an admissible solution. An example is performed in Example 2.5, see Figure 2.5(a) and 2.5(c).
(a) (b)
Fig. 2.3: The admissible solution 2-shock shock Figure 2.3(a), shock 2-shock 1-shock Figure 2.3(b)
2 We would like to remark that an admissible solution of a Riemann problem may admit a 2-shock wave which connects a left state which is on a branch of a hyperbola to a right state located on the other branch, see Figure 2.3(a). In the following section, this 2-shock will be called a non classical shock wave. It turns out that such a non classical shock wave is not easily captured by some classical numerical methods.
2.5.2. Key examples and important comments The first three Riemann problems (Example 2.3,2.4,2.5) we present do not give rise to classical weak solution made of two waves of different families while the last one (Example 2.6) produces a
non classical shock wave. The first Riemann problem gives rise two rarefactions of the 2-family, the second one leads to three waves (a 2-wave followed by and attached to a 1-wave followed by a 2-wave) and the third one produces a pure phase (α = 0) starting from a mixture and contains three shocks. It is interesting to notice that in the third Riemann problem (Example 2.5) the value of the velocity of the vanishing phase does not necessarily equal the one of the non vanishing phase. Finally, the last Riemann problem illustrates an admissible solution consisting in two waves of different families whose one is a non classical wave.
Example 2.3. The configuration αL= αR= 0.5, ωL= −ωR= 3 generates a pure gas
intermediate value U∗= (1,0). The admissible solution consists of two rarefaction waves. See Figure 2.4(a) and Figure 2.4(c).
Example 2.4. The configuration αL= 0, αR> 0, ωL> 0 > ωR is an example where
the solution is a rarefaction touching the degenerate 1-shock followed by another 2-rarefaction such that λ2(U∗) = λ2(U∗∗) = 0 and the speed of degenerate 1-shock is also
zero, where U∗= (0,0), U∗∗= (1,0) are the intermediate values. See Figure 2.4(b), 2.4(d).
Example 2.5. The configuration αL= αR= 0.5, ωL= −ωR= −5 generates a pure
liquid and the solution consists in three shocks (a 1-shock connects to 2-shock followed by another 1-shock). See Figure 2.5(a)and 2.5(c).
Example 2.6. The configuration αL= 0.8, αR= 0.5, ωL= −3, ωR= 5 generates a
non classical shock and the solution consists in two shocks (a non classical 2-shock connects to a 1-shock). See Figure 2.5(b) and 2.5(d).
(a) α, ω (b) α, ω
(c) u1, u2 (d) u1, u2
Fig. 2.4: Example 2.3: 2-rarefaction 2-rarefaction, Figure 2.4(a) and 2.4(c). Example 2.4: 2-rarefaction 1-shock 2-rarefaction, Figure 2.4(b) and 2.4(d).
(a) α, ω (b) α, ω
(c) u1, u2 (d) u1, u2
Fig. 2.5: Example 2.5: 1-shock 2-shock 1-shock, Figure 2.5(a) and 2.5(c); Example 2.6: 2-rarefaction 1-shock 2-rarefaction, Figure 2.5(b) and 2.5(d).
3. Numerical study We now investigate the numerical simulation of the sys-tem (1.4) and show that the basic Roe scheme fails to capture the expected dynamics whereas the Godunov scheme and the Roe scheme with a Harten type correction cap-ture the analytic solution. However in the non classical shock wave (corresponding to a passage through the domain H+ and H−, both of these schemes show oscillations.
We then propose a reconstruction method, see in [13],[14],[15],[16],[17] and references therein, which significantly improves the numerical result in this case.
We consider a uniform mesh of the computational domain [0,1] whose N cells are cen-tered at xi, i = 1,...,N . The space step ∆x = xi− xi−1 is constant whereas the time
step ∆t(Un) > 0 depends on the discrete field Un= (Uin)i=1,...,N which approximates
the exact solution U (x,t) at cells i and time tn=Pn−1
k=0∆t(U
k). The time step should
satisfy the following CFL condition in order to ensure the stability of the explicit schemes: ∆t ≤ ∆x
maxi{λ1(Ui,Ui+1),λ2(Ui,Ui+1)}, where λk(Ui,Ui+1) is the largest value of
|λk| on the path connecting Ui to Ui+1 using the rarefactions and admissible shock
waves computed in Theorem 2.2 and 2.3. We point out that λk(Ui,Ui+1) may be
dif-ferent from |λk(Ui)| and |λk(Ui+1)| because the characteristic fields are non genuinely
nonlinear. Denote U∗intermediate states, then
λk(Ui,Ui+1) = max
U∗ {|λk(Ui)|,|λk(Ui+1)|,|λk(U
We consider conservative finite volume schemes in the following explicit form: Uin+1= Uin−∆t ∆x Φni+1/2− Φn i−1/2 , (3.2) where Φn
i+1/2 is the numerical flux function at the interface between cells i and i + 1,
and at time tn. We compute the numerical flux Φn
i+1/2 using one of the following
strategy.
3.1. Godunov scheme
Φni+1/2= FGodj+1/2= F (U∗(Uin,Ui+1n )), where U∗(Un
i,Ui+1n )) is the value taken by the solution of the Riemann problem
between the left state Un
i and the right state Ui+1n at the interface.
3.2. Roe scheme with a Harten type correction Φni+1/2= FHarj+1/2=F (U n i ) + F (Ui+1n ) 2 − |A Roe(Un i ,U n i+1)| + har n i,i+1 Id · ∆Ui+1/2 2 , where ∆Ui+1/2= Ui+1n − U
n i , A
Roe(Un i ,U
n
i+1) is the Roe matrix, (see the Appendix
for its expression), and harni,i+1= C max(|λ1(Uin) − λ1(Ui+1n )|,|λ2(Uin) − λ2(Ui+1n )|). If
C = 0 we recover the standard Roe scheme. However it is well-known that the Roe scheme may capture non admissible solutions (see [20]). Hence we used a constant value C =1
5 to include a Harten type entropic correction in the Roe scheme.
3.3. Reconstruction scheme In [13], Lagouti`ere proposed a non-dissipative scheme which bases on an in-cell discontinuous reconstruction of the solution for scalar equations. We also refer the reader to [15],[16] for the computation of non-classical shocks with such an approach. For the system case, we refer the reader to [14],[17] and references therein for more details. One major advantage of this scheme is to capture precisely classical and non classical shock waves, a challenging point in our model owing to non classical shock waves. We do not intend to men-tion the details of the reconstrucmen-tion method. Instead, we present the main ideas of the method and summarize the computation of the numerical flux function. In particular, considering our system, the unknown variables is U = (α,ω) but only α varies in a 1-shock, whereas the 2-shock corresponds to the case where both α and ω vary at the jump. Numerical methods in general capture well the shock when only one of the two variables varies, and show difficulties in the cases where the two vari-ables vary at the jump, especially in the case of the non classical 2-shock ie when the two states located on different branches of a hyperbola. Therefore, we will de-velop the reconstruction corresponding to this specific configuration. Let us denote RP(Uj−1,Uj+1) the Riemann problem associated with the left and right initial states
Uj−1 and Uj+1. If the solution of RP(Uj−1,Uj+1) contains an admissible
disconti-nuity between the left state Uj−= (α−j (Uj−1,Uj+1),ω−j (Uj−1,Uj+1)) and right state
Uj+= (α+j (Uj−1,Uj+1),ω+j (Uj−1,Uj+1)) such that α−j 6= α + j and ω − j 6= ω + j , we propose
a discontinuous in-cell reconstruction between Uj− and Uj+in cell j as shown on Fig-ure 3.1. Otherwise, Uj−= Uj+= Uj. It is important to note that the discontinuity is
not necessarily located at the same place for both variables α and ω. We thus define the coefficient θα
j (resp. θjω) such that the distance from xj−1/2 to the discontinuity
of the variable α (resp. variable ω) is θα
• We would first like to locate the discontinuities of α and ω in a way that yields a conservative scheme. θα
j and θωj must satisfy
θα jα − j + 1 − θαj α + j = αnj, θω jω − j + 1 − θ ω j ω + j = ω n j. – If θα
j ∈ [0,1] : No reconstruction for α i.e./
α+j = α−j = αnj. – If θω
j ∈ [0.1] : No reconstruction for ω i.e./
ω+j = ωj−= ωjn. – If θα
j ∈ [0,1] (and/or θjω∈ [0,1]), reconstructing for α (and/or ω).
• Let us denote σj the exact value of the speed of propagation of the
discon-tinuity Uj−,Uj+. We then compute the numerical flux function between tn
and tn+ ∆t by using the reconstructed discontinuities rather than the
aver-age values. More precisely, if σj> 0 (resp. σj< 0), we are going to calculate
the flux at interface j + 1/2 (resp. j − 1/2) by considering that the numerical flux equals the exact flux evaluated on the right value Uj+, until the corre-sponding discontinuity reaches the interface j + 1/2 (resp. j − 1/2), and the exact flux evaluated on the left value Uj− afterwards. Therefore, such a flux function will be computed relying on the speed of shock propagation σj of
the reconstructed discontinuity and on the times ∆tαj+1/2,∆tωj+1/2needed by this discontinuity to reach the interface j ± 1/2 depending on the sign of σj.
More explicitely: – If σj> 0. Denote ∆tωj+1/2= (1−θωj)∆x σj , ∆t α j+1/2= (1−θα j)∆x σj . The
nu-merical flux function FRec
j+1/2 is computed by using U − j and U + j . ∗ If θα j ≤ θωj, then
∆tFRecj+/2= min∆tωj+1/2,∆tF (Uj+) + max∆t − ∆tαj+1/2,0F (Uj−) + maxmin∆tαj+1/2,∆t− ∆tω j+1/2,0 F (α+j,ω−j) . ∗ If θα j > θωj, then
∆tFRecj+/2= min∆tαj+1/2,∆tF (Uj+) + max∆t − ∆tωj+1/2,0F (Uj−) + maxmin∆tωj+1/2,∆t− ∆tα j+1/2,0 F (α−j,ω+j) . – If σj< 0. Denote ∆tωj−1/2= θωj∆x −σj , ∆t α j−1/2= θjα∆x −σj .
The numerical flux function FRec
j−1/2 is computed by using U − j and U + j . ∗ If θα j ≤ θωj, then
∆tFRecj−1/2= min∆tαj−1/2,∆tF (Uj−) + max∆t − ∆tωj−1/2,0F (Uj+) + maxmin∆tωj−1/2,∆t− ∆tαj−1/2,0F (α+j,ωj−) .
x α xj−1/2 xj+1/2 dαj αj−1 αj+1 αj α+j α−j x ω dωj xj−1/2 xj+1/2 ωj−1 ωj+1 ωj ω−j ωj+ Fig. 3.1: Reconstruct Uj by Uj−= (α − j,ω − j), U + j = (α + j,ω + j ) and d α j= θ α j∆x, d ω j = θωj∆x. ∗ If θα j > θωj, then
∆tFRecj−1/2= min∆tωj−1/2,∆tF (Uj−) + max∆t − ∆tαj−1/2,0F (Uj+) + maxmin∆tαj−1/2,∆t− ∆tω
j−1/2,0
F (α−j,ωj+) .
4. Numerical results We present some numerical results obtained with the constant densities ρ1= 1, ρ2= 3, which give a good overview of the wave structure.
Moreover, the simulation is implemented on a spacial domain [0,1], uniform mesh with space step ∆x and CFL number is less than or equal to 1. The time step is defined by
∆t = CFL × ∆x
maxi{λ1(Ui,Ui+1),λ2(Ui,Ui+1)}
(4.1)
where λk(Ui,Ui+1), k = 1,2 are defined by (3.1).
We first show in subsection 4.1 that the Godunov scheme and the Roe scheme with Harten type correction are able to capture the non classical wave structure joining two states in different domains H− and H+ in the Riemann problem involving a
pure phase intermediate state (Examples 2.3 and 2.5). These schemes however show strong oscillation in capturing the non classical 2-shock wave in Example 2.6, see this configuration in Figure 2.3(a), whereas the reconstructing method show very good results, Figure 4.3.
Then in subsection 4.2 we simulate the classical problem of phase separation under gravity.
4.1. The Riemann problem The Riemann problem consists in solving the system (1.4) with K = 0, S = 0 and the initial data
U (x,0) = (βL,ωL) if x ≤ 0, (βR,ωR) if x > 0.
(4.2) From Theorem 2.5, this problem admits a unique admissible solution satisfying Liu’s criterion with α1,α2∈ [0,1]. In the special case where ωL= −ωR, the solution involves
Fig. 4.1: Solution of the Riemann problem at time t = 0.15 for the initial data α1=
α2= 0.5 and ωL= −ωR= 5; 100 cells and CFL = 0.9.
a pure phase: the lighter if ωL> 0, and the heavier if ωL< 0. It consists of two
tran-sonic rarefactions in the former case and three shocks waves in the latter. We present in Figure 4.1 and 4.2, the numerical results obtained using the Godunov scheme, the Roe scheme, and the Roe scheme with the Harten type entropy fix presented at Section 3. In the first case ωL> 0, Figure 4.1, the original Roe scheme is unable to
capture the admissible solution and captures instead an inadmissible shock, i.e. does not satisfy the Liu criterion. We remark that the velocity of the liquid in the pure gas region is smooth. In the second case ωL< 0, Figure 4.2, the original Roe scheme
and others schemes capture well the pure liquid state. In this case, the gas velocity in the pure liquid region includes of three shocks and is bounded. The velocity of the vanishing phase in both of cases is not necessarily equal to the one of the pure phase. The third numerical simulation of the Riemann problem is the non classical 2-shock wave as in Figure 2.3(a). We recall that this 2-shock goes through the domain H−
and H+, connects the left state UL to the intermediate state Uint such that each
component of UL and Uint is different and the speed propagation of the 2-shock is
not equal to zero. These challenges lead to oscillations given by both the Godunov scheme and the Roe scheme with Harten entropy fix while the Roe scheme without entropy fix yields strong oscillations. The admissible solution is well captured only by the reconstruction method, see Figure 4.3 (a uniform mesh with 100 cells), Figure 4.5 (a uniform mesh with 500 cells) and Figure 4.4 for the convergence of these schemes.
Fig. 4.2: Solution of the Riemann problem at time t = 0.15 for the initial data α1=
α2= 0.5 and ωL= −ωR= −5; 100 cells and CFL = 0.9.
4.2. The phase separation under gravity This is a classical test case in the assessment of numerical methods in the modeling of counter-current two phase flows with steep transition (see [21]). We consider the model (1.1) with g = −10m/s2, K = 0
and x ∈ [0,1] with the initial data u1(x,0) = 0,u2(x,0) = 0,α1(x,0) = 0.5,α2(x,0) = 0.5
and boundary date u1(0,t) = u2(0,t) = u1(1,t) = u2(1,t) = 0. The transient result in
Figure 4.6 (left) shows that the Roe scheme captures an inadmissible shock departing from x = 1. This is consistent with the results shown in the previous section since the Riemann problems at the walls yield pure phases intermediate states and a transonic rarefaction fan for the lighter phase. However, the Roe scheme with Harten entropic correction gives a similar result to the Godunov scheme, both of them being consistent with the analysis of the Riemann problem.
Both the physical and mathematical analysis agree that the expected stationary state for the volume fraction and velocities should satisfy α1= 0 on [0,0.5] and α1= 1 on
[0.5,1] velocity u2= 0 on [0,0.5] and u1= 0 on [0.5,1]. However there is a debate as to
what should be the value of u2(resp. u1) on [0.5,1] (resp. [0,0.5]) since in that region
the liquid (resp. the gas) is absent). In our model we can compute the stationary velocity of the liquid which is not zero hence there is no mechanical equilibrium. However there does not exist a stationary value for the gas velocity on the whole of domain, we refer the reader to Appendix 5.2 for details. During the numerical simulation, all schemes except the original Roe scheme captured well the vanishing velocity of liquid in the pure gas domain as well as the (non stationary) vanishing velocity of gas in the pure liquid domain.
Fig. 4.3: Solution of the Riemann problem at time t = 0.12 for the initial data αL=
0.9,αR= 0.3 and ωL= −3,ωR= 5; 100 cells and CFL = 0.5.
5. Appendix
5.1. Appendix: model derivation We recall the two-fluid model equations for an isentropic two phase flows in one space dimension:
∂tα1ρ1+ ∂x(α1ρ1u1) = 0, (5.1a)
∂tα2ρ2+ ∂x(α2ρ2u2) = 0, (5.1b)
∂t(α1ρ1u1) + ∂x(α1ρ1u21) + α1∂xP1= α1ρ1g, (5.1c)
∂t(α2ρ2u2) + ∂x(α2ρ2u22) + α2∂xP2= α2ρ2g. (5.1d)
We assume that the two phases are incompressible (ρ1 and ρ2 are constant) and
recall the closure laws, α1+ α2= 1 and P1− P2=2(ρρ1ρ2
1−ρ2)(u1− u2)
2. Therefore the
four equation system (5.1a-5.1d) should be solved for the four unknowns (α1, u1, u2,
and P1).
Applying the same method as in [2], we derive a system of two equations, which allows for the study of the void waves and avoids singularities when one phase disappears. From the equations of mass, (5.1a) and (5.1b), we obtain
∂x(α1u1+ α2u2) = 0. (5.2)
Therefore the quantity
Fig. 4.4: Solution of the Riemann problem at time t = 0.12 for the initial data αL=
0.9,αR= 0.3 and ωL= −3,ωR= 5; 500 cells and CFL = 0.5.
is constant in space and can be determined from the boundary conditions. Assuming that the boundary condition is independent of time, we obtain that K is constant both in space and time.
From the equation of momentum conservation (5.1c) and (5.1d), we derive
u1ρ1(∂tα1+ ∂xα1u1) + α1ρ1(∂tu1+ u1∂xu1) + α1∂xP1= α1ρ1g, (5.4a)
u2ρ2(∂tα2+ ∂xα2u2) + α2ρ2(∂tu2+ u2∂xu2) + α2∂xP2= α2ρ2g. (5.4b)
Thank to (5.1a) and (5.1b), these equations can be simplified:
α1 ∂t(ρ1u1) + ∂x ρ1u21 2 + α1∂xP1= α1ρ1g, (5.5a) α2 ∂t(ρ2u2) + ∂x ρ2u22 2 + α2∂xP2= α2ρ2g. (5.5b)
Assuming that initially α1α26= 0, we can simplify by α1in (5.5a) and by α2in (5.5b),
then subtracting the two equations yields
∂t(ρ1u1− ρ2u2) + ∂x 1 2(ρ1u 2 1− ρ2u22) + P1− P2 = (ρ1− ρ2)g. (5.6)
Fig. 4.5: Convergence curves (mesh refinement for the Riemann problem with initial data αL= 0.9,αR= 0.3 and ωL= −3,ωR= 5; CFL = 0.5).
(a) Transient (b) Stationary
Fig. 4.6: Volume fraction α1for the sedimentation problem.
The space differential in (5.6) can be simplified thanks to 1 2(ρ1u 2 1− ρ2u22) + P1− P2= 1 2(ρ1u 2 1− ρ2u22) + ρ1ρ2 2(ρ1− ρ2) (u1− u2)2 = 1 2(ρ1− ρ2) (ρ1u1− ρ2u2)2. (5.7)
We set the new unknowns (α,ω) as α = α1,
ω = ρ1u1− ρ2u2. (5.8)
The original unknowns u1 and u2 can be recovered from (5.3) and (5.8):
u1= (1 − α)ω α(ρ2− ρ1) + ρ1 + Kρ2 α(ρ2− ρ1) + ρ1 , u2= −αω α(ρ2− ρ1) + ρ1 + Kρ1 α(ρ2− ρ1) + ρ1 .
(a) Transient (b) Stationary
Fig. 4.7: The velocity u1 for the sedimentation problem.
(a) Transient (b) Stationary
Fig. 4.8: The velocity u2 for the sedimentation problem.
Finally, from (5.1a), (5.6) and (5.7) we obtain the 2 × 2 system ∂tα + ∂xαu1= 0, ∂t(ω + (ρ1− ρ2)K) + ∂x (ω+(ρ 1−ρ2)K)2 2(ρ1−ρ2) = (ρ1− ρ2)g,
and the incompressible two-fluid model can be written in closed form as ∂tα + ∂x α(1−α)ω α(ρ2−ρ1)+ρ1+ Kα = 0, ∂tω + ∂x ω2 2(ρ1−ρ2)+ Kω = (ρ1− ρ2)g, (5.9)
since K is constant in space and time.
5.2. Appendix: Vanishing velocity of the sedimentation problem We consider the stationary state of the 2 × 2 system (5.9) assuming that g < 0 and ρ1< ρ2.
with (5.3) yield
α1u1= 0,
α2u2= 0.
We seek a solution consisting of two zones. A bottom zone with pure phase 2: α2= 1
and constant velocity u2= 0 in the region x ∈ [0,0.5] and a top zone with pure phase
1: α1= 1 and constant velocity u1= 0 in the region x ∈ [0.5,1]. We are going to use
the second equation of the system (5.9) to determine the vanishing velocity of phase 2 in the region x ∈ [0.5,1].
The second equation of the system (5.9), using the physical variables u1 and u2 (or
equivalently equation 5.6) is ∂x ρ1 u2 1 2 − ρ2 u2 2 2 + ρ1ρ2 2(ρ1− ρ2) (u1− u2)2 = (ρ1− ρ2)g. (5.10)
Integrating (5.10) we obtain the two phase Bernoulli’s principle:
ρ1 u2 1 2 − ρ2 u2 2 2 + ρ1ρ2 2(ρ1− ρ2) (u1− u2)2− (ρ1− ρ2)gx = constant. (5.11)
In order to compute the vanishing phase velocity of phase 2, we remark that the velocities at the walls x = 1 are u1= u2= 0. Hence the constant in (5.11) equals −(ρ1−
ρ2)g, and since u1= 0 for x ∈ [0.5,1] the two phase Bernouilli’s principle becomes
ρ22 ρ1− ρ2 u22(x) 2 = (ρ1− ρ2)g(x − 1) for x ∈ [0.5,1]. Hence u2(x) = − s 2 1 −ρ1 ρ2 g(x − 1) for x ∈ [0.5,1]. (5.12)
The velocity profile is therefore not constant and furthermore shows a discontinuity at the interface x = 0.5.
We remark that we cannot determine a vanishing velocities for phase 1 in the re-gion x ∈ [0,0.5]. Indeed since u2= 0 for x ∈ [0,0.5], the two-phase Bernouilli’s principle
takes the form
ρ2 1 ρ1− ρ2 u2 1(x) 2 = (ρ1− ρ2)gx for x ∈ [0,0.5]. (5.13) Since g = −10m/s < 0, the equation (5.13) has no solution as it yields u2
1< 0.
5.3. Appendix: The Roe matrix A Roe matrix A(UL,UR) for the system
(1.4) and two states UL,UR∈ H is a diagonalisable matrix such that
F (UL) − F (UR) = A(UL,UR)(UL− UR)
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